To solve the problem step by step, we will break it down into two parts as given in the question.
### Part (a): Find the number of photons passing through unit area per second at a distance of 2 m from the source.
1. **Understanding Power and Energy of Photons**:
- The power \( P \) of the source is given as \( 10 \, \text{W} \).
- Power is defined as energy per unit time, so \( P = \frac{E}{t} \).
- Each photon has energy given by the formula \( E = \frac{hc}{\lambda} \), where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m} \).
2. **Calculating Energy of One Photon**:
\[
E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{589 \times 10^{-9} \, \text{m}}
\]
\[
E \approx 3.37 \times 10^{-19} \, \text{J}
\]
3. **Calculating Number of Photons Emitted per Second**:
- The number of photons emitted per second \( n_0 \) can be calculated as:
\[
n_0 = \frac{P}{E} = \frac{10 \, \text{W}}{3.37 \times 10^{-19} \, \text{J}} \approx 2.97 \times 10^{19} \, \text{photons/s}
\]
4. **Finding the Area of a Sphere**:
- At a distance \( r = 2 \, \text{m} \), the area \( A \) of a sphere is given by:
\[
A = 4\pi r^2 = 4\pi (2^2) = 16\pi \, \text{m}^2
\]
5. **Calculating Number of Photons per Unit Area**:
- The number of photons passing through unit area per second \( n \) is given by:
\[
n = \frac{n_0}{A} = \frac{2.97 \times 10^{19}}{16\pi} \approx 5.91 \times 10^{17} \, \text{photons/m}^2/\text{s}
\]
### Part (b): Calculate the distance where the mean concentration of photons is \( 100 \, \text{cm}^{-3} \).
1. **Convert Concentration to SI Units**:
- The mean concentration \( n \) is given as \( 100 \, \text{cm}^{-3} = 100 \times 10^6 \, \text{m}^{-3} \).
2. **Volume of a Spherical Shell**:
- The volume \( V \) of a spherical shell with radius \( r \) and thickness \( dr \) is given by:
\[
V = 4\pi r^2 dr
\]
3. **Photons Crossing the Shell**:
- The number of photons crossing this shell per unit time is:
\[
n \cdot V \cdot \frac{dr}{dt} = n \cdot 4\pi r^2 \cdot c
\]
4. **Setting Up the Equation**:
- The number of photons emitted per second \( n_0 \) is equal to the number crossing the shell:
\[
n_0 = n \cdot 4\pi r^2 \cdot c
\]
- Substitute \( n_0 = \frac{10 \lambda}{hc} \):
\[
\frac{10 \lambda}{hc} = n \cdot 4\pi r^2 \cdot c
\]
5. **Solving for \( r \)**:
- Rearranging gives:
\[
r^2 = \frac{10 \lambda}{4\pi n h c^2}
\]
- Substitute \( \lambda = 589 \times 10^{-9} \, \text{m} \), \( n = 100 \times 10^6 \, \text{m}^{-3} \), \( h = 6.626 \times 10^{-34} \, \text{J s} \), and \( c = 3 \times 10^8 \, \text{m/s} \):
\[
r^2 = \frac{10 \times 589 \times 10^{-9}}{4\pi \times (100 \times 10^6) \times (6.626 \times 10^{-34}) \times (3 \times 10^8)^2}
\]
- Calculate \( r \) to find \( r \approx 8.88 \, \text{m} \).
### Final Answers:
- (a) The number of photons passing through unit area per second at a distance of 2 m is approximately \( 5.91 \times 10^{17} \, \text{photons/m}^2/\text{s} \).
- (b) The distance where the mean concentration of photons is \( 100 \, \text{cm}^{-3} \) is approximately \( 8.88 \, \text{m} \).
